voidquick_sort(int q[], int l, int r) { if (l >= r) return ; int i = l - 1, j = r + 1, x = q[l + r >> 1]; while (i < j) { do i ++ ; while (q[i] < x); do j -- ; while (q[j] > x); if (i < j) swap(q[i], q[j]); } quick_sort(q, l, j); quick_sort(q, j + 1, r); }
intmain() { int n; scanf("%d", &n); for (int i = 0; i < n; i ++ ) scanf("%d", &q[i]); quick_sort(q, 0, n - 1); for (int i = 0; i < n; i ++ ) printf("%d ", q[i]); return0; }
#include<iostream> usingnamespace std; constint N = 1e5 + 10;
int a[N], tmp[N]; voidmerge_sort(int q[], int l, int r) { if (l >= r) return ; int mid = l + r >> 1; merge_sort(q, l, mid), merge_sort(q, mid + 1, r); int k = 0, i = l, j = mid + 1; while (i <= mid && j <= r) if (q[i] <= q[j]) tmp[k ++ ] = q[i ++ ]; else tmp[k ++ ] = q[j ++ ]; while (i <= mid) tmp[k ++ ] = q[i ++ ]; while (j <= r) tmp[k ++ ] = q[j ++ ]; for (int i = l, j = 0; i <= r; i ++ , j ++ ) q[i] = tmp[j]; }
intmain() { int n; scanf("%d", &n); for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]); merge_sort(a, 0, n - 1); for (int i = 0; i < n; i ++ ) printf("%d ", a[i]); return0; }
逆序对的数量
给定一个长度为 n 的整数数列,请你计算数列中的逆序对的数量。
逆序对的定义如下:对于数列的第 i 个和第 j 个元素,如果满足 i<j 且 a[i]>a[j],则其为一个逆序对;否则不是。